Question 66

In $$\triangle ABC, \angle B = 68^\circ$$ and $$\angle C = 32^\circ$$. Sides $$AB$$ and $$AC$$ are produced to points $$D$$ and $$E$$, respectively. The bisectors of $$\angle DBC and \angle BCE$$ meet at F. What is the measure of $$\angle BFC$$?

Solution

From the angle bisector property,

$$\angle BFC = 90\degree - \frac{\angle A}{2}$$

In $$triangle ABC$$,

$$\angle A + \angle B + \angle C = 180\degree$$

$$\angle A = 180 - 32 - 68 = 80\degree$$

$$\angle BFC = 90\degree - \frac{80\degree}{2}$$

$$\angle BFC = 50\degree$$

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